forcesfree-body-diagramfrictionnewtonian-mechanics
First of all sorry for this basic question.
Newton's third law states that for every force there is a force, which is equal in magnitude and opposite in direction to those of given force.So by that logic friction should also have a reaction force.
Now my questions are, what is the reaction force for static friction ?
For instance in the picture, what is the direction of reaction force of static friction and on which object will that force act ?
Static friction always opposes relative motion at the point of contact.
There are two cases possible:
1)It orients itself in direction and magnitude in such a way that the relative acceleration of the contact point is zero.
2)If this is not possible(such as in friction is too small to prevent motion),it tries to minimize the relative acceleration.
Firstly, are these steps correct? Will I always obtain the correct answer by following these steps?
Yes and yes, those steps are correct :)
Are there some other steps to follow when I know that the object is definitely accelerating in one direction or another?
I don't think so. Let's work through your example using those steps:
To stop the block slipping downwards, the frictional force must be equal and opposite to the downwards force \begin{align} \vec{Fr}_{BA} &= - \vec{W_B} \\ \mu |\vec{N_{BA}}| \hat{y} &= -(-mg)\hat{y} = mg\hat{y} \end{align} If you consider the horizontal components you can see that $|\vec{N_{BA}}| = m|\vec{a}|$, so $$ |\vec{a}| = \frac{g}{\mu} = 19.62 \ \text{ms}^{-2} $$
To stop A slipping upwards past B, the frictional force on A must be equal and opposite to the upwards force \begin{align} \vec{Fr_{AB}} &= - \vec{P_{vert}} \\ &= -(-W_B) \\ -\mu |\vec{N_{AB}}| \hat{y} &= -(--mg)\hat{y} = -mg\hat{y} \end{align} If you consider the horizontal components you can see that $|\vec{N_{AB}}| = m|\vec{a}|$, so $$ |\vec{a}| = \frac{g}{\mu} = 19.62 \ \text{ms}^{-2} $$ which is also the same as in the inertial frame... nailed it! :) Now if you're like me, you're probably thinking "that was the worst piece of nasty, hateful, misery I've ever seen." And you'd be right! My advice would be just pick an inertial frame and be done with it.
Let's walk through a more simple example; just block A and the table (no block B).
Take the vector-sum of all forces on A to calculate the resultant force on A, $\vec{R_A}$ \begin{align} \vec{R_A} &= \vec{D} + \vec{N_{AT}} + \vec{W_A} \\ &= D \hat{x} + (N_{AT} - W_A) \hat{y} \\ &= D \hat{x} + 0 \hat{y} \end{align} Now we resolve $\vec{R_A}$ into components that are parallel and perpendicular to the slope \begin{align} \vec{R}_{parallel} &= D \hat{x} \\ \vec{R}_{perpendicular} &= 0 \hat{y} \end{align}
We know the direction of static friction is opposite to $\vec{R}_{parallel}$, so $\hat {Fr_{A}} = -\hat{x} $. Done :)
The resultant force on the table is the vector sum of all forces on it \begin{align} \vec{R_{T}} &= \vec{P_{horz}} + \vec{N_{TG}} + \vec{W_T} + \vec{W_A} + \vec{N_{AG}} \\ &= -P_{horz} \hat{x} + (N_{TG} - W_T + W_A - N_{AG})\hat{y} \\ &= -D \hat{x} + 0 \hat{y} \end{align} Resolving this into components: \begin{align} \vec{R}_{T, \ parallel} &= -D \hat{x} \\ \vec{R}_{T, \ perpendicular} &= 0 \hat{y} \end{align}
The direction of static friction on the table is opposite $\vec{R}_{T, \ parallel}$:
$$\vec{Fr}_{TA} = -\vec{R}_{T, \ perpendicular} = -(-D) \hat{x} = D \hat{x}$$
Every action has an equal and opposite reaction, so as A is exerting a frictional force on the table rightwards, the table must be exerting a frictional force leftwards on A and voila
$$
\hat{Fr}_{AT} = -\hat{Fr}_{TA} = -\hat{x}
$$
To be honest, pseudo forces make my brain explode. There's a helpful example over on wikipedia. In the example, the person is analogous to block B and the seat is analogous to block A. If you imagine being block B it's like being accelerated in a car; you get pushed back into your seat (which is the pseudo-force). Then at constant velocity you feel like your not moving, but the road is moving under you.
It's always worth bearing in mind these useful rules:
I'll leave what happens in A's frame as a fun exercise for you.
Best Answer
There is no reaction force on this free body diagram. That is because this is the diagram for the object only.
Draw a free body diagram of the Earth, and there you have your reaction force. You will see that the same static friction pulls the opposite way in the Earth.
Free-body force diagrams tell a story about one object (or system) and are thus useful for e.g. Newton's 2nd and 1st laws. But the action/reaction force couples that come from Newton's third law appear between objects, or you could say that they appear on two different objects. Not on the same objects. Therefor they might not both be included within a free-body diagram - if they are both included (when you define your system as including both objects), then they will always exactly cancel out and will thus make no different for Newton's 2nd and 1st laws, and thus we ignore them. You will thus never see an action/reaction force pair both appearing on a free-body diagram.
To explain your specific scenario: The static friction force is trying to prevent sliding from happening. So it pulls leftwards in the box in order to try to prevent it from moving, and it pulls rightwards in the Earth to try to make the Earth follow along with the box, so that there is no sliding.